Initial

1.org

@@ -0,0 +1,60 @@
+#+title: Assignment 1
+#+options: toc:nil
+#+latex_header: \usepackage{parskip}
+
+In order to pass this assignment you have to get at least two points.
+
+(Exercises marked with [BN] are based on exercises from a previous version of this course, given by Bengt Nordström.)
+
+* (1 point)
+State and give a brief explanation of the Church-Turing thesis.
+
+** Answer
+The Church-Turing thesis states that "effective" computation can be carried out by several methods.
+In this case, effective means a set of structured finite instructions, that requires no knowledge except for the instructions themselves.
+Further, given that the instructions, given that no error occurs, then the result should be completed in finite time.
+* (1 point)
+Give an example of a function on the natural numbers (from $\mathbb{N}$ to $\mathbb{N}$) that is:
+- Injective but not surjective.
+- Bijective.
+
+** Answer
+$\lambda x. x + 1$ is injective and not surjective. This comes from that each natural number has a successor (injective), but $0$ is not the image of any value (not surjective).
+
+$\lambda x. x$ is bijective, which is trivial, since each input is their own image.
+
+* (1 point)
+Is $\mathbb{N} \times \text{Bool}$ countable? Provide a proof. (Bool is a set with two elements, true and false.)
+
+** Answer
+$\mathbb{N} \times \text{Bool}$ is countable.
+One way to prove this is by isomorphism between $\mathbb{N} \times \text{Bool}$ and $\mathbb{N} + \mathbb{N}$.
+To show this isomorphism, we need two functions $f \in \mathbb{N} \times \text{Bool} \rightarrow \mathbb{N} + \mathbb{N}$ and $g \in \mathbb{N} + \mathbb{N} \rightarrow \mathbb{N} \times \text{Bool}$, such that $f \circ g = \text{Id}_{\mathbb{N} + \mathbb{N}}$ and that $g \circ f = \text{Id}_{\mathbb{N} \times \text{Bool}}$.
+So, let $f = \lambda (n,b).\ \text{if}\ b\ \text{then}\ \text{Inr}\ n\ \text{else}\ \text{Inl}\ n$ and $g = \lambda (x).\ \text{case}\ x\ \text{of}\ \text{Inl}\ n \rightarrow (n,\text{False});\ \text{Inr}\ n \rightarrow (n,True)$
+Now, these are trivially isomorphic, and thus will not be shown.
+
+Since this these functions are isomorphic, they are also injective, which means that if $\mathbb{N} + \mathbb{N}$ is countable, then $\mathbb{N} \times \text{Bool}$ will also be so.
+In the literature for this week (namely Enumerable sets and Hilbert's hotel), it is shown that $\mathbb{N} + \mathbb{N}$ is countable, and therefore $\mathbb{N} \times \text{Bool}$.
+
+*** Alternative solution
+
+This can be answered by creating a function $f \in \mathbb{N} \times \text{Bool} \rightarrow \mathbb{N}$ that is injective.
+I define it as $f = \lambda (x,b). if b then 2 \cdot x + 1 else 2 \cdot x$.
+To show that it is injective, let's do a proof by contradiction.
+Assume that $(n_1, b_1) \ne (n_2, b_2)$ and $f (n_1, b_1) = f (n_2, b_2) = c$.
+If $c$ is odd, then $b_1$ and $b_2$ have be $\text{True}$, further, then $n_1$ and $n_2$ have to $\frac{c - 1}{2}$, thus creating a contradiction.
+If $c$ is even, then $b_1$ and $b_2$ have be $\text{False}$, further, then $n_1$ and $n_2$ have to $\frac{c}{2}$, thus creating a contradiction.
+In both cases a contradiction is raised, meaning that the function is injective, and thus $\mathbb{N} \times \text{Bool}$ is countable.
+
+* (1 Point) [BN]
+Is $\mathbb{N} \rightarrow \text{Bool}$ countable? Provide a proof.
+
+** Answer
+No.
+
+Assume that it is countable.
+Then one could enumerate them as $f_0, f_1, \dots$.
+Then lets create a function $d = \lambda i. \neg (f_i i)$.
+Assume that $d$ is in the set of functions, that is to say that $d = f_k$ for some k.
+If that was the case, then $d\ k = \neg (f_k\ k) = \neg (d\ k)$ which is a contradition.
+Thus $d$ cannot be in the set of functions, which in of it self creates a contradiction that it is enumerable.